In a series LCR circuit, $C = 80 μF, L = 5.0 H$ and $R = 40 Ω$ are connected to a variable frequency source of 240 V. The angular frequency of the source which drives the circuit at resonance will be
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → $50\, rad\, s^{-1}$
Given:
$L = 5.0\ H$, $C = 80\ \mu F = 80 \times 10^{-6}\ F$
At resonance, the angular frequency is
$\omega = \frac{1}{\sqrt{LC}}$
Substituting the values:
$\omega = \frac{1}{\sqrt{5 \times 80 \times 10^{-6}}}$
$\omega = \frac{1}{\sqrt{4 \times 10^{-4}}}$
$\omega = \frac{1}{2 \times 10^{-2}} = 50\ rad/s$
∴ The angular frequency at resonance is $50\ rad/s$.